Question: You have found the following ages (in years) of all 5 snakes at your local zoo: $ 22,\enspace 20,\enspace 21,\enspace 13,\enspace 41$ What is the average age of the snakes at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 snakes at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{22 + 20 + 21 + 13 + 41}{{5}} = {23.4\text{ years old}} $ Find the squared deviations from the mean for each snake. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $22$ years $-1.4$ years $1.96$ years $^2$ $20$ years $-3.4$ years $11.56$ years $^2$ $21$ years $-2.4$ years $5.76$ years $^2$ $13$ years $-10.4$ years $108.16$ years $^2$ $41$ years $17.6$ years $309.76$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{1.96} + {11.56} + {5.76} + {108.16} + {309.76}} {{5}} $ $ {\sigma^2} = \dfrac{{437.2}}{{5}} = {87.44\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{87.44\text{ years}^2}} = {9.4\text{ years}} $ The average snake at the zoo is 23.4 years old. There is a standard deviation of 9.4 years.